The energy of one mole of photons of radiation of wavelength $$300$$ nm is (Given :
$$h = 6.63 \times 10^{-34}$$ J s, $$N_A = 6.02 \times 10^{23}$$ mol$$^{-1}$$, $$c = 3 \times 10^{8}$$ m s$$^{-1}$$)

We need to find the energy of one mole of photons of radiation with wavelength $$\lambda = 300$$ nm; here $$h = 6.63 \times 10^{-34}$$ J s, $$N_A = 6.02 \times 10^{23}$$ mol$$^{-1}$$, and $$c = 3 \times 10^8$$ m/s.

The energy of a single photon is given by:

$$E = \frac{hc}{\lambda}$$

Since the wavelength is provided in nanometres, we convert:

$$\lambda = 300 \text{ nm} = 300 \times 10^{-9} \text{ m} = 3 \times 10^{-7} \text{ m}$$

Substituting into the formula leads to:

$$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3 \times 10^{-7}}$$

This simplifies to:

$$E = \frac{19.89 \times 10^{-26}}{3 \times 10^{-7}}$$

and so:

$$E = 6.63 \times 10^{-19} \text{ J}$$

Now, multiplying by Avogadro's number gives the energy per mole of photons:

$$E_{mole} = N_A \times E = 6.02 \times 10^{23} \times 6.63 \times 10^{-19} \text{ J}$$

From this we obtain:

$$E_{mole} = 6.02 \times 6.63 \times 10^{23-19} \text{ J}$$

which yields:

$$E_{mole} = 39.9126 \times 10^{4} \text{ J}$$

and therefore:

$$E_{mole} = 3.99 \times 10^{5} \text{ J} = 399 \text{ kJ/mol}$$

Therefore, the correct option is Option C: 399 kJ mol$$^{-1}$$.